Consider a rocket that is in deep space and at rest relative to an inertial reference frame. The rocket’s engine is to be fired for a certain interval. What must be the rocket’s mass ratio (ratio of initial to final mass) over that interval if the rocket’s original speed relative to the inertial frame is to be equal to (a) the exhaust speed (speed of the exhaust products relative to the rocket) and (b)2.0times the exhaust speed?

The initial velocity of rocket,$V_i=0\mathrm{m}/\mathrm{s}$

Here, we can use the second rocket equation to calculate the mass ratio in both cases.

Formula:

$v_f-v_i=v_{rel}\times \mathrm{In}\left(\frac{M_i}{M_f}\right)$

We have, the second rocket equation as,

$v_f-v_i=v_{rel}\times \mathrm{In}\left(\frac{M_i}{M_f}\right)$

Here,$v_f$is the original speed of rocket relative to the inertial frame of reference.

And$v_{rel}$is the exhaust speed of rocket

So, substituting the value of$v_i$and rearranging above equation for mass ratio, we get

$\frac{M_i}{M_f}=\exp \left(\frac{{\mathrm{v}}_{\mathrm{f}}}{{\mathrm{v}}_{\mathrm{rel}}}\right)$

When,$v_f=2\times v_{rel}$ the above equation become,

role="math" localid="1661250600364" $$\begin{gathered} \frac{M_i}{M_f}=\exp \left(2\right) \\ \cong 7.4 \end{gathered}$$

Hence, the rocket’s mass ratio when the rocket’s original speed is equal to the exhaust speed$\frac{M_i}{M_f=7.4}$

When $v_f=2\times v_{rel}$the above equation become,

$$\begin{gathered} \frac{M_i}{M_f}=\exp \left(2\right) \\ \cong 7.4 \end{gathered}$$

Hence, the rocket’s mass ratio when the rocket’s original speed is times the exhaust speed, $\frac{M_i}{M_f}=7.4$